Sum of Reciprocals of Powers as Euler Product/Proof 2

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Theorem

Let $\zeta$ be the Riemann zeta function.

Let $s\in \C$ be a complex number with real part $\sigma>1$.


Then $\zeta(s) = \displaystyle\prod_p\frac1{1-p^{-s}}$ where the infinite product runs over the prime numbers.


Proof

From Sum of Geometric Progression:

$\dfrac 1 {1 - p^{-z} } = 1 + \dfrac 1 {p^z} + \dfrac 1 {p^{2 z} } + \cdots$


From Sum of Reciprocals of Powers is Absolutely Convergent iff Modulus of Power is Greater than One:

$\displaystyle \sum_{n \mathop \ge 1} n^{-z}$ is absolutely convergent

if and only if:

$\left\lvert{z}\right\rvert \ge 1$

Thus:

\(\displaystyle \sum_p \dfrac 1 {1 - p^{-z} }\) \(=\) \(\displaystyle \sum_p \left({1 + \dfrac 1 {p^z} + \dfrac 1 {p^{2 z} } + \dfrac 1 {p^{3 z} } + \cdots}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({1 + \dfrac 1 {2^z} + \dfrac 1 {2^{2 z} } + \dfrac 1 {2^{3 z} } + \cdots}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(\) \(\, \displaystyle \times \, \) \(\displaystyle \left({1 + \dfrac 1 {3^z} + \dfrac 1 {3^{2 z} } + \dfrac 1 {3^{3 z} } + \cdots}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(\) \(\, \displaystyle \times \, \) \(\displaystyle \left({1 + \dfrac 1 {5^z} + \dfrac 1 {5^{2 z} } + \dfrac 1 {5^{3 z} } + \cdots}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(\) \(\, \displaystyle \times \, \) \(\displaystyle \cdots\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 1 + \dfrac 1 {2^z} + \dfrac 1 {3^z} + \dfrac 1 {2^{2 z} } + \dfrac 1 {5^z}\) $\quad$ $\quad$
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \dfrac 1 {2^z 3^z} + \dfrac 1 {7^z} + \dfrac 1 {2^{3 z} } + \dfrac 1 {3^{2 z} }\) $\quad$ $\quad$
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \cdots\) $\quad$ $\quad$

The result follows from the Fundamental Theorem of Arithmetic.

$\blacksquare$


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